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jek_recluse [69]
3 years ago
6

Major species present when fructose is dissolved in water

Chemistry
2 answers:
Katyanochek1 [597]3 years ago
7 0
The fructose chemical formula is C6H12O6. The answer to the question above regarding the major species present when fructose is dissolved in water (H2O) is "None". No ions are present. It is false that when sugar is dissolved in water there will be strong electrolytes.
BARSIC [14]3 years ago
6 0

Answer:

Alfa-fructofuranose and beta-fructofuranose

Explanation:

The species more stables in water of fructose are their cyclic forms. In aqueous solution, the ketone group of the fructose reacts in a reversible way with the hydroxyl group of the same forming a cyclic hemiketal, in a reaction of intramolecular cycling. These rings formed are more stables when contain 5 or 6 atoms, named furanose and pyranose forms, respectively. In this cycling process the carbonyl carbon (C=O) it is transformed into a chiral center which produces 2 possible anomers, alfa and beta. In the fructose, the hydroxyl group of the formed hemiketal it is produced in the carbon 2 (the anomeric carbon) and can be located above the ring or beneath the ring. In D-sugars (the most common in nature), when the hydroxyl group is located beneath the ring the structure is in the alfa form and when is above the ring is in the beta form.

In the case of fructose dissolved in water it is formed the alfa-fructofuranose an beta-fructofuranose in a majority way, as shown in the picture.

Sugars are capable of suffering of mutarotation when they are dissolved in water, which cause the interconversion of anomers alfa and beta, producing a mix of the alfa and beta forms in the furanose or pyranose rings. And the proportion of each

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1. If a solution containing 48.99 g of mercury(II) perchlorate is allowed to react completely with a solution containing 8.564 g
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1. Chemical eqn:

Hg(ClO4)2 + Na2S -> HgS + 2NaClO4

mercury perchlorate has a molar mass of 399.5g/mol (1d.p) and for Na2S it is 78.0g/mol (1d.p.)

no. of moles of mercury perchlorate= 48.99÷399.5= 0.12263mol(5s.f.)

no. of moles of Na2S= 8.564÷78.0= 0.10979mol ( 5s.f.)

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so no. of moles of Hg(ClO4)2 needed if all Na2S is used up is 1/1×0.10979= 0.10979 mols

since no. of moles of mercury perchlorate needed < no. of moles of it provided, it is in excess and Na2S is the limiting factor.

since HgS is a solid and NaClO4 is aqueous, solid ppt formed will only be HgS.

no. of moles of HgS/ no. of moles of NaClO4= 1/1

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molar mass of HgS= 232.7g/mol 1d.p.

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no. of excess moles= 0.12263-0.10979= 0.01284mols

grams of excess reactant= 0.01284×399.5= 5.13g (3s.f.)

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